For the given circuit the current through battery of 6 V just after closing the switch ' S ' will be…

Just after closing the switch $S$, inductor behaves like an open circuit, so no current will flow through $2 \mathrm{\Omega }$ resistors.

Thus, current through the circuit is $I=\frac{6}{2+4}=1 \mathrm{A}$ (Here $2 \mathrm{\Omega }$ and $4 \mathrm{\Omega }$ resistors are in series).